Need help doing this simple differentiation.
Consider 4 d Euclidean(or Minkowskian) spacetime.
\begin{equation}
\partial_{\mu}\frac{(a-x)_\mu}{(a-x)^4}= ?
\end{equation}
where $a_\mu$ is a constant vector and the indices are summed over since one really doesn't need to bother about upper and lower indices in flat space.
Also $(a-x)^2=(a-x)_\mu(a-x)_\mu$. A simple minded calculation gives me the result $0$. But I think the answer may contain Dirac Delta function like the relation below in 3 dimension
\begin{equation}
{\bf{\nabla}} . \frac{\hat{r}}{r^2}=4\pi \delta ^3(r)
\end{equation}
A corollary question. What will be $\partial_{\mu}\frac{1}{(a-x)^2}$ ? Is it
$\frac{2(a-x)_\mu}{(a-x)^4}$ or does it also involve the Delta function?

2 Answers
2

In a $d$-dimensional Euclidean space (with positive definite norm), one has
$$ \vec{\nabla} \cdot \frac{\vec{r}}{r^d}
~=~{\rm Vol}(S^{d-1})~\delta^d(\vec{r}), $$
cf. the divergence theorem and arguments involving either test functions and integration by part, or $\epsilon$-regularization, similar to methods applied in this Phys.SE answer. Here ${\rm Vol}(S^{d-1})$ is the surface area of the $(d-1)$-dimensional units sphere $S^{d-1}$.

By similar arguments one may show that the identity
$$ \vec{\nabla}(r^{2-d}) ~=~(2-d)\frac{\vec{r}}{r^d} , \qquad d\neq 2, $$
contains no distributional contributions in $d$-dimensional Euclidean space.

For the related questions in Minkowski space, one suggestion is to introduce an $\epsilon$-regularization in the Euclidean formulation, and then perform a Wick rotation, and at the end of the calculation, let $\epsilon\to 0^+$.

Thanks Qmechanic. Writing down your equation in the index notation gives
\begin{equation}
\partial_\mu \frac{x_\mu}{x^2}=2\pi^2\delta^4(x)
\end{equation}
in 4d which is the result I wanted. Now it looks a bit silly to have asked the question in the first place.